Q17-75E - Organic Chemistry (Mcmurry)

Question

Compound A, ${\mathbf{C}}_8{\mathbf{H}}_{\mathbf{1}0}\mathbf{O}$ , has the IR and ${}^{\mathbf{1}}\mathbf{H}$ NMR spectra shown. Proposea structure consistent with the observed spectra, and label each peak inthe NMR spectrum. Note that the absorption at $\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{\delta }$ disappears when ${\mathbf{D}}_{\mathbf{2}}\mathbf{O}$ is added.

Step-by-Step Solution

Answer

Step 1: Given formula

The formula ${\text{C}}_8{\mathrm{H}}_{10}\mathrm{O}$ shows that the carbon present in molecule are 8 and we know in alkane the number of hydrogen are twice the number of carbon plus 2. The number of hydrogen here would have been 18 in case of alkane. So the multiple bonds present are 4. Indicating ring may also be present.

Step 2: Table for IR absorption by specific functional group

Given table has values for absorption of radiation at particular wavelength by specific functional group:

                                             Functional group absorption band

From table and molecular formula it can be concluded:

• The broad peak at indicates a hydroxyl group.
• The peak at $1500\text{ and 1600 }{\mathrm{cm}}^{-1}$ must be due to aromatic ring.
• The absorption at $\text{830 }{\mathrm{cm}}^{-1}$ shows ring is $\mathrm{p}\text{ - disubstituted.}$

Step 3: 1 H NMR spectrum reading

Single peak indicates no neighbor H atom is present to hydrogen. If peak splits into two it indicates only one neighbor H is present and if into three then two neighbor H are  present and if four peaks then four neighbor hydrogens and so on.

Step 4: Table for chemical shift in NMR

Given table has values forchemical shift:

                                                            Chemical shift

From table, molecular formula and IR spectrum, it can be concluded:

• The chemical shift  indicates presence of 3H (primary alkyl) and the chemical shift   indicates presence of 2H (secondary alkyl) attached to aromatic ring. Both peaks are for the ethyl group present.

 

• The two peakaround chemical shift $6.7-7$ are due to aromatic or aryl protons. The symmetrical splitting shows that the ring is $\mathrm{p}\text{ - disubstituted.}$ 

 

• The chemical shift of  is due to an OH proton.

Step 5: Structure of given molecule

From above data it can be concluded that below given structure is correct structure of given molecule, ${\mathrm{C}}_8{\mathrm{H}}_{10}\mathrm{O}$

Step 6: Peak labeling

In graph:

• The peak at indicates ${\mathrm{CH}}_3$  $$\begin{gathered} {\mathrm{CH}}_3 \\ {\mathrm{C}}_8{\mathrm{H}}_{10}{\mathrm{O}}_2 \\ 3.76\mathrm{\delta } \end{gathered}$$and peak at 2.55 indicates ${\mathrm{CH}}_2$ attached to aromatic ring. Both peaks are for the ethyl group that is present at aromatic ring.

 

• The peak around $6.7-7$ are due to aromatic or aryl protons so show aromatic ring.

 

The peak at 5.5 shows OH group